![Ex 3.2, 12 - Given 3 [x y z w] = [x 6 -1 2w] + [4 x+y z+w 3]](https://d1avenlh0i1xmr.cloudfront.net/acde432f-1f3a-494e-9eee-a021e5848322/slide40.jpg)
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Ex 3.2, 12 - Chapter 3 Class 12 Matrices (Term 1)
Last updated at Jan. 17, 2020 by Teachoo
Ex 3.2
Ex 3.2, 1
Ex 3.2, 2 (i)
Ex 3.2, 2 (ii) Important
Ex 3.2, 2 (iii)
Ex 3.2, 2 (iv)
Ex 3.2, 3 (i)
Ex 3.2, 3 (ii) Important
Ex 3.2, 3 (iii)
Ex 3.2, 3 (iv) Important
Ex 3.2, 3 (v)
Ex 3.2, 3 (vi) Important
Ex 3.2, 4
Ex 3.2, 5
Ex 3.2, 6
Ex 3.2, 7 (i)
Ex 3.2, 7 (ii) Important
Ex 3.2, 8
Ex 3.2, 9
Ex 3.2, 10
Ex 3.2, 11
Ex 3.2, 12 Important You are here
Ex 3.2, 13 Important
Ex 3.2, 14
Ex 3.2, 15
Ex 3.2, 16 Important
Ex 3.2, 17 Important
Ex 3.2, 18
Ex 3.2, 19 Important
Ex 3.2, 20 Important
Ex 3.2, 21 (MCQ) Important
Ex 3.2, 22 (MCQ) Important
Chapter 3 Class 12 Matrices
Serial order wise
Transcript
Ex 3.2, 12 Given 3[■8(x&y@z&w)] = [■8(x&6@−1&2w)] + [■8(4&x+y@z+w&3)] find the values of x, y, z and w. 3[■8(x&y@z&w)] = [■8(x&6@−1&2w)] + [■8(4&x+y@z+w&3)] [■8(3x&3y@3z&3w)] = [■8(x+4&6+x+y@1−z+w&2w+3)] Since matrices are equal. Corresponding elements are equal 3x = x + 4 3y = 6 + x + y 3z = 1 – z + w 3w = 2w + 3 …(1) …(2) Solving equation (1) 3x = x + 4 3x – x = 4 2x = 4 x = 4/2 x = 2 Solving equation (2) 3y = 6 + x + y 3y – y = 6 + x 2y = 6 + x Putting x = 2 2y = 6 + 2 2y = 8 y = 8/2 y = 4 Solving equation (4) 3w = 2w + 3 3w – 2w = 3 w = 3 Solving equation (3) 3z = – 1 + z + w 3z – z = – 1 + w 2z = – 1 + w Putting w = 3 2z = – 1 + 3 2z = 2 z = 2/2 z = 1 Hence, x = 2, y = 4 , w = 3 & z = 1
